Problem: $f(x, y) = x - 3y$ We have a change of variables: $\begin{aligned} x &= X_1(u, v) = \dfrac{u}{5} + \dfrac{v}{4} \\ \\ y &= X_2(u, v) = \dfrac{u}{2} - \dfrac{v}{2} \end{aligned}$ What is $f(x, y)$ under the change of variables? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{-12u}{7} + \dfrac{12v}{5}$ (Choice B) B $\dfrac{-9u}{8} + \dfrac{5v}{4}$ (Choice C) C $\dfrac{-13u}{10} + \dfrac{7v}{4}$ (Choice D) D $\dfrac{-7u}{5} + \dfrac{4v}{3}$
When applying a change of variables, we substitute the new definition for $x$ and $y$ into the original equation. The original equation: $f(x, y) = x - 3y$ Let's substitute $X_1(u, v)$ for $x$ and $X_2(u, v)$ for $y$. $\begin{aligned} f(x, y) &= \left( \dfrac{u}{5} + \dfrac{v}{4} \right) - 3 \left( \dfrac{u}{2} - \dfrac{v}{2} \right) \\ \\ &= \dfrac{u}{5} + \dfrac{v}{4} - \dfrac{3u}{2} + \dfrac{3v}{2} \\ \\ &= \dfrac{4u}{20} + \dfrac{5v}{20} - \dfrac{30u}{20} + \dfrac{30v}{20} \\ \\ &= \dfrac{-26u}{20} + \dfrac{35v}{20} \\ \\ &= \dfrac{-13u}{10} + \dfrac{7v}{4} \end{aligned}$ Therefore, under the change of variables, $f(x, y)$ becomes: $\dfrac{-13u}{10} + \dfrac{7v}{4}$